Question: $f(x) = \dfrac{ \sqrt{ x - 1 } }{ x^2 + 13 x + 36 }$ What is the domain of the real-valued function $f(x)$ ?
Answer: $f(x) = \dfrac{ \sqrt{ x - 1 } }{ x^2 + 13 x + 36 } = \dfrac{ \sqrt{ x - 1 } }{ ( x + 9 )( x + 4 ) }$ First, we need to consider that $f(x)$ is undefined anywhere where the radical is undefined, so the radicand (the expression under the radical) cannot be less than zero. So $x - 1 \geq 0$ , which means $x \geq 1$ Next, we also need to consider that $f(x)$ is undefined anywhere where the denominator is zero. So $x \neq -9$ and $x \neq -4$ However, these last two restrictions are irrelevant since $1 > -9$ and $1 > -4$ and so $x \geq 1$ will ensure that $x \neq -9$ and $x \neq -4$ Combining these restrictions, then, leaves us with simply $x \geq 1$ Expressing this mathematically, the domain is $\{ \, x \in \RR \mid x \geq1\, \}$.